Question

A motor bike initially at rest picks up a velocity 72km/hr over a distance of 40m find acceleration and time in which it picks up velocity

Answer 1

Answer:

5 m/s² & 4 s

Explanation:

Use equation of motion

• v² - u² = 2aS

a = (v² - u²)/(2S)

= [(72 × 5/18 m/s)² - (0 m/s)²] / (2 × 40 m)

= (400/80) m/s²

= 5 m/s²

Now, from equation of motion

• v = u + at

t = (v - u)/a

= [(72 × 5/18 m/s) - (0 m/s)] / (5 m/s²)

= (20/5) s

= 4 s

Answer 2

$\large\blue{heya mate!!}$

v² - u² = 2aS

a = (v² - u²)/(2S)

= [(72 × 5/18 m/s)² - (0 m/s)²] / (2 × 40 m)

= (400/80) m/s²

= 5 m/s²

Now, from equation of motion

v = u + at

t = (v - u)/a

= [(72 × 5/18 m/s) - (0 m/s)] / (5 m/s²)

= (20/5) s

= 4 s